从后台上传图片,用getcwd()得到文件路径,如"E:/SOURCE/SERVICE/UPLOADIMAGE",但是存入数据库后就变成"E:SOURCESERVIECEUPLOADIMAGE",中间的/没了,导致取不出正确的路径,哪位大神可以帮忙解决一下问题啊
$upload_dir = getcwd();
$newfilename = makefilename();
$oldfilename = $_FILES['upfile']['name']; // 服务器端临时文件名
$pos = strrpos($oldfilename, '.'); // 获取文件名中最右侧的.
$ext = substr($oldfilename, $pos, strlen($oldfilename)-$pos); // 获取文件的扩展名
$newspic = $upload_dir. $newfilename . $ext;
$up_name=$_FILES['upfile']['name'];//客户端文件名
代码如上,需要存入数据库的为$newspic
$upload_dir = getcwd();
$newfilename = makefilename();
$oldfilename = $_FILES['upfile']['name']; // 服务器端临时文件名
$pos = strrpos($oldfilename, '.'); // 获取文件名中最右侧的.
$ext = substr($oldfilename, $pos, strlen($oldfilename)-$pos); // 获取文件的扩展名
$newspic = $upload_dir. $newfilename . $ext;
$up_name=$_FILES['upfile']['name'];//客户端文件名
代码如上,需要存入数据库的为$newspic
$in_sql="insert into upfile(newsid,upfilename,up_name,upfiledis,upfileurl,uptime,new_upfilename)values('$newsid','$upfilename','$up_name','$upfiledis','$upfileurl','$uptime','$newfilename') ";
if(mysql_query($in_sql)){
?>
<script>alert('恭喜您,添加成功!');window.location='new_display.php?typeid=<?php echo $typeid?>&borderid=<?php echo $borderid?>'; </script>";
<?PHP
}else{
?>
<script>alert('很抱歉,添加失败!');window.location='new_display.php?typeid=<?php echo $typeid?>&borderid=<?php echo $borderid?>';
?>
<script>alert('恭喜您,添加成功!');window.location='new_display.php?typeid=<?php echo $typeid?>&borderid=<?php echo $borderid?>'; </script>";
<?PHP
}else{
?>
<script>alert('很抱歉,添加失败!');window.location='new_display.php?typeid=<?php echo $typeid?>&borderid=<?php echo $borderid?>'; </script>";